1. In the adjacent figure, two triangles are similar find the length of the missing side
Solution:
Let the triangles be ABC and PQR
BC/QR = AC/PR
5/X = 13/39
13X = 5 X 39
X = 5X39/13 = 5 X3 = 15
- What number is to 12 is 5 is 30?
Solution
Let x be the number
x:12 :: 5 : 30
30x = 12×5
x = 12×5/30 = 2
- Solve the following properties:
(i). x : 5 = 3 : 6
(ii) 4 : y = 16 : 20
(iii) 2 : 3 = y : 9
(iv) 13 : 2 = 6.5 : x
(v) 2 : π = x : 22/7
Solution:
(i). x : 5 = 3 : 6
6x = 5 x 3
6x = 15
x = 15/6
(ii) 4 : y = 16 : 20
4×20 = 16y
y = 4×20/16
y = 5
(iii) 2 : 3 = y : 9
2×9 = 3y
y = 2×9/3 = 2×3 = 6
(iv) 13 : 2 = 6.5 : x
13x = 2 x 6.5
13x = 13
x = 13/13 = 1
(v) 2 : π = x : 22/7
2x22/7 = πx
x = (2x22/7) /π =(2x22/7) /(22/7)
x = 2
- find the mean proportion to :
(i) 8, 16
(ii) 0.3, 2.7
(ii)162/3 , 6
(iv) 1.25, 0.45
Solution:
(i) 8, 16
Let x be the mean proportion to 8 and 16
Then 8/x = x/16
x2 = 8 x 16 = 128
x = √128 = √(64×2) = 8√2
(ii) 0.3, 2.7
Let x be the mean proportion to 0.3 and 2.7
Then 0.3/x = x/2.7
x2 = 0.3 x 2.7 = 0.81
x = √(0.81) = 0.9
(ii)162/3 , 6
Let x be the mean proportion to 162/3 and 6
Then (162/3) /x = x/6
x2 = 162/3 x 6 = 50/3 x 6 = 100
x = √100 = 10
(iv) 1.25, 0.45
Let x be the mean proportion to 1.25 and 0.45
Then 1.25/x = x/0.45
x2 = 1.25 x 0.45
x = √(1.25 x 0.45) = √(1.25 x 0.45)x√(100×100)/ √(100×100)
= √(125×45)/√(100×100) = √(25x5x5x9)/√(100×100) = 5x5x3/10×10 = 3/4
- Find the fourth proportion for the following:
(i) 2.8, 14, 3.5
(ii) 31/3, 12/3, 21/2
(iii)15/7, 23/4, 33/5
Solution:
(i) 2.8, 14, 3.5
Let x be the fourth proportion
Then, 2.8 : 14 :: 3.5 : x
2.8x = 14×3.5
x = 14×3.5/2.8 = 17.5
(ii) 31/3, 12/3, 21/2
Let x be the fourth proportion
Then, 31/3: 12/3 :: 21/2: x
10/3 :5/3 : : 5/2 : x
10/3x = 5/3 x 5/2
10/3x = 25/6
x = 25/6 x 3/10 = 75/60 = 15/12 = 5/4
(iii)15/7, 23/14, 33/5
Let x be the fourth proportion
Then, 15/7: 23/14:: 33/5: x
12/7 :31/14 : : 18/5 : x
12/7 x = 31/14 x 18/5
12/7 x = 31×18/14×5
x = 31×18/14×5 x 7/12 = 31×3/5×4 = 93/20 = 413/20
- Find the third proportion to:
(i) 12, 16
(ii) 4.5, 6
(iii) 51/2 , 161/2
Solution:
(i) 12, 16
Let x be the third proportion
Then 16:12 :: x : 16
12x = 16 x16
x = 16×16/12 = 64/3 = 211/3
(ii) 4.5, 6
Let x be the third proportion
Then 6:4.5 :: x : 6
4.5x = 6 x6
x = 6×6/4.5 = 36/4.5 = 360/45 = 8
(iii) 51/2 , 161/2
Let x be the third proportion
Then 161/2: 51/2:: x : 161/2
11/2 x = 33/2 x 33/2
x = 33/2 x 33/2 x 2/11= 33×3/2 = 99/2 = 491/2
- In a map 1/4 cm represents 25km, if two cities are 21/2c apart on the map, what is the actual distance between them?
Solution:
Let 21/2 cm represemts x km
1/4cm: 25km :: 21/2cm : x km
1/4 x x = 25 x 21/2
x/4 = 25 x5 /2
x = 25×5/2 x 4 = 25x5x2 = 250km
- Suppose 30out of 500 components for a compter were found defective. At this aret how many defective components would he found in 1600 components?
Solution:
Number of defective components in 500 components = 30
Let x be the number of defecrive components in 1600 components
then 30:500 :: x :1600
30×1600 = 500x
x = 30×1600/500 =96
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